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I'm trying to show that given a set $\{\mathbf{a}, \mathbf{b}\}$ of orthonormal vectors in a 2-dimensional vector space, I can construct the identity matrix by computing $aa^\dagger + bb^\dagger$. This should be straightforward but it's not working out. I get that my conditions for orthonormality are $$|a_1|^2+|a_2|^2 = 1,$$ $$|b_1|^2 + |b_2|^2 = 1,$$ $$a_1^*b_1+a_2^*b_2 = 0$$ but these don't directly lead me to the identity matrix. Where am I going wrong?

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If you are over $\mathbb C$, shouldn't it read $aa^* + bb^*$? – martini Sep 21 '12 at 13:17
Is your vector space complex? Otherwise the stars in your last equations don't seem to make sense. But if it is complex, then the transposes in $aa^T+bb^T$ should rather be adjoints. – Henning Makholm Sep 21 '12 at 13:18
@MTurgeon: From context it seems clear that the vectors are rows. martini's point is that complex conjugates should be used if working over $\mathbb C$, which seems consistent with the last equation (where $a_1^*$ and $a_2^*$ appear). – Jonas Meyer Sep 21 '12 at 13:20
I am using the convention that vectors are columns and this is a complex vector space. I'm also letting my components be $a_1$ and $a_2$ be components for $\mathbf{a}$ etc – Yardl Sep 21 '12 at 13:30
@JonasMeyer I see. I completely misunderstood the question then. – M Turgeon Sep 21 '12 at 13:39

As $a$ and $b$ are orthonormal, they form a basis of your vector space $V$. We have \[ (aa^* + bb^* )a = aa^* a + bb^* a = a|a|^2 + b0 = a \] and \[(aa^* + bb^* )b = aa^* b + bb^* b = a0 + b|b|^2 = b \] So $(aa^* + bb^*)$ is the identity on a basis, hence on $V$, which gives $aa^* + bb^* = \mathrm{Id}$.

$\def\abs#1{\left|#1\right|}$ More explicitly we first exploit $a_1^*b_1 + a_2^*b_2 = 0$, which gives \[ \abs{a_1}^2\abs{b_1}^2 = \abs{a_1^*b_1}^2 = \abs{a_2^*b_2}^2 = \abs{a_2}^2\abs{b_2}^2 \] Now $\abs{b_1}^2 = 1 - \abs{b_2}^2$ and $\abs{a_2}^2 = 1- \abs{a_1}^2$, that is \[ \abs{a_1}^2 - \abs{a_1}^2\abs{b_2}^2 = \abs{b_2}^2 - \abs{a_1}^2\abs{b_2}^2\iff \abs{a_1} = \abs{b_2} \] and therefore also $\abs{a_2} = \abs{b_1}$.

So we have \begin{align*} a_1^*a_1 + b_1^*b_1 &= \abs{a_1}^2 + \abs{b_1}^2\\ &= \abs{a_1}^2 + \abs{a_2}^2\\ &= 1,\\ a_2^*a_2 + b_2^*b_2 &= \abs{a_2}^2 + \abs{b_2}^2\\ &= \abs{b_1}^2 + \abs{b_2}^2\\ &= 1 \end{align*} and hence the elements on the diagonal of $aa^* + bb^*$ are 1, as wished. It remains to conside the off-diagonal elements. We have \begin{align*} a_1(a_1^*a_2 + b_1^*b_2) &= \abs{a_1}^2a_2 + a_1b_1^*b_2\\ &= \abs{b_2}^2a_2 + a_1b_1^*b_2\\ &= b_2b_2^*a_2 + a_1b_1^*b_2\\ &= (b_1^*a_1 + b_2^*a_2)b_2\\ &= 0. \end{align*} So either $a_1 = 0$ or $a_1^*a_2 + b_1^*b_2 = 0$. In the letter case, we are done, in the former case, $a_2 \ne 0$ (as $\abs{a_2}^2 + \abs{a_1}^2 = 1$) and hence $a_2^* \ne 0$, but \begin{align*} a_2^*(a_1^*a_2 + b_1^*b_2) &= \abs{a_2}^2a_1^* + a_2^*b_1^*b_2\\ &= \abs{b_1}^2a_1^* + a_2^*b_1^*b_2\\ &= b_1^*b_1a_1^* + a_2^*b_1^*b_2\\ &= (a_1^*b_1 + a_2^*b_2)b_1^*\\ &= 0. \end{align*} So we are done in this case also.

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This is fine but why doesn't explicit matrix multiplication work out? – Yardl Sep 21 '12 at 13:43
It does, see above. – martini Sep 21 '12 at 14:02
Awesome, thanks. I just thought that it would somehow show up immediately but apparently it requires some more work. – Yardl Sep 21 '12 at 14:13

Denote the two vectors by $\{\mathbf{a},\mathbf{b}\}$ instead. Then orthonormality is the condition that $\mathbf{a}^\dagger\mathbf{b}=\mathbf{a}^\dagger\mathbf{b}=0$, while $\mathbf{a}^\dagger\mathbf{a}= \mathbf{b}^\dagger\mathbf{b}=1$. Now $\mathbf{a}\mathbf{a}^\dagger + \mathbf{b}\mathbf{b}^\dagger$ is the block matrix product $$\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right),$$ where each factor is a $2\times 2$ matrix. This product is the identity matrix iff the two factors are inverses, which we can check by evaluating the product in the other order: $$\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right) = \left(\begin{array}{cc}\mathbf{a}^\dagger\mathbf{a} & \mathbf{a}^\dagger\mathbf{b}\\ \mathbf{b}^\dagger\mathbf{a} & \mathbf{b}^\dagger\mathbf{b}\end{array}\right)=\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right).$$ So $\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)$ and $\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)$ are indeed inverses, and $$\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)=\mathbf{a}\mathbf{a}^\dagger + \mathbf{b}\mathbf{b}^\dagger$$ is the $2\times 2$ identity matrix.

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